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Vlad
3 months ago
13

Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in

still water and the rate of the current.

Physics
2 answers:
Sav [3.1K]3 months ago
8 0

Representing the boat's speed as

v_b = 10 mph

and noting that the river's speed is given as

v_r = 2 mph

Explanation:

When the boat travels downstream, its effective speed is determined by combining the boat's speed with the river's current since they move in the same direction:

v_1 = v_r + v_b

Conversely, when the boat moves upstream, its net speed is reduced by the river's current, resulting in:

v_2 = v_b - v_r

As we observe, the boat covers 24 miles downstream in 2 hours:

v_1 = \frac{24}{2} = 12 mph

Similarly, for the upstream journey, it covers the same distance in 3 hours:

v_2 = \frac{24}{3} = 8 mph

v_b + v_r = 12 mph

v_b - v_r = 8 mph

Therefore, the speed of the boat is given by

v_b = 10 mph

and the river speed is described as

v_r = 2 mph

Softa [3K]3 months ago
3 0

The boat's speed in still water is 10 mph.

The current's speed is 2 mph.

Further explanation

Acceleration is defined as the rate at which velocity changes.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

a = acceleration (m/s²), v = final velocity (m/s)

u = initial velocity (m/s)

t = duration (s)

d = distance (m)

Let's delve into this problem!

Given:

Distance d = 24 miles

Time taken to travel downstream = td = 2 hours

Time taken for the upstream return = tu = 3 hours

Unknowns:

Speed of the boat in still water = vs =?

Speed of the current = vc =?

Solution:

While Marcus sails downstream, the boat's speed aligns with the current's velocity.

v_s + v_c = \frac{d}{t_d}

v_s + v_c = \frac{24}{2}

v_s + v_c = 12

v_s = 12 - v_c → Equation 1

When proceeding upstream, the boat's speed counters the current's force.

v_s - v_c = \frac{d}{t_d}

v_s - v_c = \frac{24}{3}

v_s - v_c = 8

12 - v_c - v_c = 8 ← Equation 1

12 - 2v_c = 8

2v_c = 12 - 8

2v_c = 4

v_c = 4 \div 2

\large {\boxed {v_c = 2 ~ mph} }

v_s = 12 - v_c

v_s = 12 - 2

\large {\boxed {v_s = 10 ~ mph} }

Learn more

  • Velocity of Runner:
  • Kinetic Energy:
  • Acceleration:
  • The Speed of Car:

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity, Driver, Car, Deceleration, Acceleration, Obstacle, Speed, Time, Rate

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